a 95 kg man lying on a surface of negligible friction shoves a 74 g stone away from himself, giving it a speed of 4.0 m/s. what speed does the man acquire as a result?



Answer :

ayune

As a result of the law of  conservation of momentum, the man will acquire a speed of 3 .  10⁻³ m/s

The law of conservation of momentum states thta:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Where:

u₁ , u₂ = initial velocity of object 1 and object 2

v₁ , v₂ = final velocity of object 1 and object 2

Since both the man and the shove are initially at rest, hence u₁ , u₂ = 0.

Therefore,

m₁v₁ + m₂v₂ = 0

Parameters given:

m₁ = 95 kg

m₂ = 74 gram =  74 . 10⁻³ kg

v₂ = 4 m/s

Plug these parameters into the equation:

95 . v₁ + 74 . 10⁻³ . 4 = 0

v₁ = 3 .  10⁻³ m/s

Hence, the man acquire a speed of 3 .  10⁻³ m/s, which is very small.

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